Hypergeometric solutions of some algebraic equations^{∗}

A.M. Perelomov ^{†}

Max-Planck-Institut für Mathematik, Vivatsgasse 7, D-53111 Bonn, Germany

To the memory of A.N. Tyurin

Abstract

We give the hypergeometric solutions of some algebraic equations including the general fifth degree equation.

1. It is well known that the general algebraic equation of degree n ≥ 5 cannot be solved by radicales [2]. However, it may be solved if we use wider classes of functions. For example, the fifth degree equation may be solved in modular functions [3], [4], [5] and as series [6]. The general algebraic equation may be solved in hyperelliptic theta constants [7].

Another approach consists in the consideration of the algebraic solutions of the differential equations and was used in the classical Schwarz’s paper [8], where all algebraic solutions of the standard hypergeometric equation were found. It was intensively developed by Poincaré.

The results of Schwarz were completed in the paper [9] by the classification of the algebraic generalized hypergeometric functions _{n}F_{n−1}. There is one series corresponding to the case under consideration, and the 76 exceptional cases. Note that for n ≥ 9 there are no exceptional cases.

∗ This paper aroused from the problem of normalization of the ground state wave function of the trigonometric n-particle Calogero-Sutherland system (see the review paper [1]). The relation of this problem with the results of the present note will be considered later.

†On leave of absence from Institute for Theoretical and Experimental Physics, 117259 Moscow, Russia. Current e-mail address: perelomo@dftuz.unizar.es

In [9] the monodromy groups for these functions were calculated explicitly. They are the groups generated by the complex reflections in ℂ^{n}. The differential Galois groups of the corresponding hypergeometric differential equations are also explicitly calculated there. However, the corresponding algebraic equations were not calculated.

In this note we give the explicit solution of the equations of type

F_{n}(x, t) ≡ x^{n} − x + t = 0,  n > 1,  (1)
G_{n}(y, t) ≡ γ_{n}t^{n−1}y^{n} − (y − 1) (y + \frac{1}{n − 1})^{n−1} = 0,  γ_{n} = \frac{n^{n}}{(n − 1)^{n−1}}

(2)

in the class of hypergeometric functions of type

n−1F_{n−2}(α_{1}, …, α_{n−1}; β_{1}, …, β_{n−2}; γt^{n−1}).  (3)

Note that the general equations of degrees 2, 3, 4 and 5 may be reduced to equation (1) by Tschirnhaus transformation, see [10]^{1}.

2. First, let us remark that the normalization of the ground state wave function of the n-particle Calogero-Sutherland problem is equivalent to the calculation of the integral

∫_{0}^{π} ... ∫_{0}^{π} { ∏_{1≤j<l≤n} sin(q_{j} − q_{l})}^{k} dq_{1} ⋯ dq_{n}.  (4)

In the process, there appear binomial coefficients of special type

c_{k}^{(n,j)} = ( \begin{matrix}nk + j \\ k \end{matrix} ) = \frac{(nk + j)!}{((n − 1)k + j)! k!} .  (5)

It is natural to introduce for them the generating functions

H_{n,j}(z) = ∑_{k=0}^{∞} c_{k}^{(n,j)}z^{k},  n ≥ 2, j = 0, 1, ….  (6)

1 After this work had been completed, Prof. Glasser kindly informed me on his paper [11], where it is shown that the roots of equation (1) are finite sums of generalized hypergeometric functions. I learned also from Prof. Zagier that the solution of this equation is known to him, unpublished.

Let us introduce also the functions

y_{n,j}(t) = t^{j} H_{n,j}(z),  where z = t^{n−1}, j = 0, 1, ….  (7)
x_{n,j}(t) = j ∫_{0}^{t} y_{n,j−1}(τ) dτ,  j = 1, 2, ….  (8)

These functions have a number of nice properties, in particular, x_{n,1}(t) and y_{n,0}(t) are the solutions of equations (1) and (2).

Recall the definition of the generalized hypergeometric function:

mF_{n}(α_{1}, α_{2}, …, α_{m}; β_{1}, β_{2}, … β_{n}; γz) = ∑_{k=0}^{∞} c_{k}z^{k} =

∑_{k=0}^{∞} \frac{(α_{1})_{k} (α_{2})_{k} ⋯ (α_{m})_{k}}{(β_{1})_{k} (β_{2})_{k} ⋯ (β_{n})_{k} k!} (γz)^{k},  (9)

where

(α)_{k} = α(α + 1) ⋯ (α + k − 1)

is the Pochhammer symbol.

Lemma 1. The functions (6)–(8) are algebraic and of hypergeometric type. Namely,

H_{n,j}(z) = _{n−1}F_{n−2} (\frac{j + 1}{n}, …, \frac{j + n}{n}; \frac{j + 1}{n − 1}, …, \frac{j + n − 1}{n − 1}; γ_{n}z) , (10)

where

γ_{n} = \frac{n^{n}}{(n − 1)^{n−1}} .  (11)

The functions y_{n,j}(t) are given by formula (7). We have also

x_{n,j}(t) =

t^{j} _{n−1}F_{n−2} (\frac{j}{n}, \frac{j + 1}{n}, …, \frac{j + n − 1}{n}; \frac{j + 1}{n − 1}, \frac{j + 2}{n − 1}, …, \frac{j + n − 1}{n − 1}; γ_{n}z) .

(12)

Proof. This follows from the fact that for the generalized hypergeometric function (9)

\frac{c_{k+1}}{c_{k}} = \frac{(k + α_{1})(k + α_{2}) ⋯ (k + α_{m})}{(k + β_{1})(k + β_{2}) ⋯ (k + β_{n})} γ,  (13)

and we have the identities

nF_{n−1}(α_{1}, …, α_{n−1}, α; β_{1}, …, β_{n−2}, α; z) =
n−1F_{n−2}(α_{1}, …, α_{n−1}; β_{1}, …, β_{n−2}; z).  (14)

The algebraic character of the functions H_{n,j}(z) becomes evident after the comparison of (10) with the list of functions from [9].

The hypergeometric character of these functions gives the possibility to easily differentiate and integrate them. For this we may use the well-known formulae:

\frac{d}{dz} _{m}F_{n}(α_{1}, …, α_{m}; β_{1}, …, β_{n}; z) =

\frac{α_{1} ⋯ α_{m}}{β_{1} ⋯ β_{n}} _{m}F_{n}(α_{1} + 1, …, α_{m} + 1; β_{1} + 1, …, β_{n} + 1; z), (15)

∫ _{m}F_{n}(α_{1} ⋯ α_{m}; β_{1} ⋯ β_{n}; z) dz =

\frac{(β_{1} − 1) ⋯ (β_{n} − 1)}{(α_{1} − 1) ⋯ (α_{m} − 1)} _{m}F_{n}(α_{1} − 1, ⋯, α_{m} − 1; β_{1} − 1, ⋯, β_{n} − 1; z) + const.

(16)

From these formulae we obtain

\frac{d}{dz} H_{n,j}(z) =

(j + n)_{n−1}F_{n−2} (\frac{j + n + 1}{n}, …, \frac{j + 2n}{n}; \frac{j + n}{n − 1}, …, \frac{j + 2n − 2}{n − 1}; γ_{n}z)

(17)

and formula (12).

3. Note that equation (1) has the solution x(t), x(0) = 0, analytic in the disc D_{r_{n}} of some radius r_{n} > 0. Consider also the function y(t) = x′(t). Differentiating (1) with respect to t, we obtain

(1 − nx^{n−1}) y = 1.  (18)

Equation (1) yields also

x^{n−1} = \frac{1}{n} (1 − \frac{1}{y}) .  (19)

From this it follows equation (2).

The main result of this paper is

Theorem 1. The solutions of equations (1) and (2) are given by the formulae

x(t) = x_{n,1}(t),  y(t) = y_{n,0}(t).  (20)

We also have

(x(t))^{j} = x_{n,j}(t),  (x(t))^{j} y(t) = y_{n,j}(t).  (21)

4. Before to prove this theorem, we give some properties of generalized hypergeometric functions (for more details see [12] and [13]).

The generalized hypergeometric function _{n}F_{n−1}(z) is defined by the series

nF_{n−1}(α_{1}, …, α_{n}; β_{1}, …, β_{n−1}; z) = ∑_{k=0}^{∞} \frac{(α_{1})_{k} ⋯ (α_{n})_{k}}{(β_{1})_{k} ⋯ (β_{n−1})_{k} k!} z^{k}  (22)

which converges in the unit disc.

The function u(z) = _{n}F_{n−1}(z) satisfies the hypergeometric equation

D · (D + β_{1} − 1) ⋯ (D + β_{n−1} − 1)u − z (D + α_{1}) ⋯ (D + α_{n}) u = 0 (23)

with D = z d/dz. This function is analytic on the Riemann sphere ℂP^{1} = ℂ^{1} ∪ {∞} except the points z = 0, z = ∞ and z = 1. The local behaviour of this function at these points is defined by the exponents

\begin{matrix}1 − β_{1}, …, 1 − β_{n−1}, 0 & at z = 0, \\ α_{1}, …, α_{n} & at z = ∞, \\ 0, 1, …, n − 2, γ = ∑_{j=1}^{n−1} β_{j} − ∑_{l=1}^{n} α_{l} & at z = 1. \end{matrix}  (24)

If the numbers β_{1}, …, β_{n−1} are distinct mod ℤ, then n independent solutions of equation (21) are given by the formula

z^{1−β_{i}} _{n}F_{n−1}(1 + α_{1} − β_{i}, …, 1 + α_{n} − β_{i}; 1 + β_{1} − β_{i}, ˆ…, 1 + β_{n} − β_{i}; z), (25)

where i = 1, …, n, β_{n} = 1, and the sign ˆ… denotes that the argument 1 + β_{i} − β_{i} is omitted.

In [12] the following proposition was proved.

Lemma 2. [12]. If in (24) γ ∉ ℕ then near the point z = 1 equation (23) has n − 1 analytic solutions of the form

u_{j}(z) = (z − 1)^{j−1} + O ((z − 1)^{n−1})  (26)

for j = 1, …, n − 1 corresponding to the exponents 0, 1, …, n − 2. Now we are ready to prove

Theorem 2. The function y(t) = x′(t) has the following properties:

1. y(t) is analytic in the disc

D_{r_{n}} = \{t: |t| < r_{n}\},  where r_{n}^{n−1} = \frac{(n − 1)^{n−1}}{n^{n}} ,  (27)

and the expansion of y(t) as t → 0 is of the form

y(t) = ỹ(z) = 1 + c_{1}z + c_{2} z^{2} + ⋯,  where z = t^{n−1}. (28)

2. The function ỹ(z) is analytic on ℂP^{1} and its values are distinct from zero and infinity except at points z = z_{0} = r_{n}^{n−1} = γ_{n}^{−1} and z = ∞. At z = z_{0} it is equal to infinity, and at z = ∞ it vanishes.

As |t| → ∞, we have

x(t) = t ∑_{k=1}^{n−1} z^{−α_{k}}g_{k}(z), ỹ(z) = ∑_{k=1}^{n−1} z^{−α_{k}} g̃_{k}(z), α_{k} = \frac{k}{n}, z = t^{n−1}.

(29)

At z = ∞, the functions g_{k}(z) and g̃_{k}(z) are analytic and their values are distinct from zero and infinity.

3. Near the point z = z_{0}, the function ỹ(z) has the form

ỹ(z) = (z − z_{0})^{−1/2}f_{1}(z) + f_{2}(z),  (30)

where the functions f_{1}(z) and f_{2}(z) are analytic at z = z_{0} with f_{1}(z_{0}) ≠ 0 and f_{2}(z_{0}) ≠ ∞.

4. The dimension of the linear space of all branches of the function ỹ(z) obtained by going around the branch points z = z_{0} and z = ∞ is equal to n − 1.

Proof.

1. The first statement of theorem follows from formulae (18), (19) and from the solution of equation (1) by iterations

x = t + x^{n},  (31)
x = t + (t + x^{n})^{n},  x = t + (t + (t + x^{n})^{n})^{n} , …,  (32)
x = t + \frac{c_{1}}{n} t^{n} + \frac{c_{2}}{2n − 1} t^{2n−1} + \frac{c_{3}}{3n − 2} t^{3n−2} + ⋯,  (33)
c_{1} = n,  c_{2} = \frac{(2n)(2n − 1)}{1 · 2},  c_{3} = \frac{3n (3n − 1)(3n − 2)}{1 · 2 · 3}, · · · . (34)

The radius of convergence of this series is determined by the condition that the discriminant of equation (1) vanishes. From this we derive the formula

z_{0} = r_{n}^{n−1} = γ_{n}^{−1} = \frac{(n − 1)^{n−1}}{n^{n}} .  (35)

2. This follows from the iterations of formula

x(t) = ε_{n} t^{1/n} (1 − \frac{x}{t})^{1/n} ,  where ε_{n} = exp (i \frac{π}{n}) ,  as |t| → ∞.

(36)

3. In the same way, we obtain the expansion of ỹ(z) at the point z = z_{0}.

4. Equation (1) has n solutions x^{(j)}(t), j = 1, …, n, x^{(1)}(t) = x(t), and we have ∑ x^{(j)}(t) = 0.

Therefore, the n branches ỹ^{(1)}(z), …, ỹ^{(n)}(z) of the algebraic function ỹ(z) = x′(t) are linearly dependent

ỹ^{(1)}(z) + ỹ^{(2)}(z) + ⋯ + ỹ^{(n)}(z) = 0,  (37)

and the dimension of the linear space of all branches of the function ỹ(z) is equal to n − 1.

Now we are ready to prove Theorem 1.

We give two different proofs of this theorem. The first proof is based on the general idea of a beautiful Riemann’s paper [14] adapted to the problem under consideration. The second one uses the standard calculation of the residue.

Proof 1. From equations (1) and (2) it follows that the function ỹ(z) = (1 − nx^{n−1})^{−1}, z = t^{n−1}, vanishes only at z = ∞, and it is equal to infinity only at z = z_{0}. Let ỹ^{(1)}(z), …, ỹ^{(n)}(z) be all the n branches of this function. Then

ỹ^{(1)}(z) + ỹ^{(2)}(z) + ⋯ + ỹ^{(n)}(z) = 0,  (38)

and the space L_{n} of linear combinations of such functions is (n−1)-dimensional. At the points z = 0, z = z_{0} and z = ∞, the expansion of the function f(z) ∈ L_{n} is defined by the exponents:

α_{1} = \frac{1}{n}, …, α_{n−1} = \frac{n − 1}{n},  at z = ∞,
β_{1} = \frac{1}{n − 1}, …, β_{n−2} = \frac{n − 2}{n − 1},  at z = 0,
γ_{1} = 0, γ_{2} = 1, …, γ_{n−2} = n − 3, γ_{n−1} = − \frac{1}{2}  at z = z_{0}.

On the other hand, n − 1 linearly independent solutions, u_{1}(z), … u_{n−1}(z), of the hypergeometric equation

D·(D+β_{1}−1) ⋯ (D+β_{n−2}−1) u(z)−z(D+α_{1}) ⋯ (D+α_{n−1}) u(z) = 0 (39)

with D = z d/dz have the same exponents and satisfy the conditions of the Riemann theorem. Therefore, according to this theorem

ỹ(z) = ỹ^{(1)}(z) ∈ L_{n},  ỹ(z) = c_{1} u_{1}(z) + ⋯ + c_{n−1} u_{n−1}(z).  (40)

The constants c_{k} may be found from the consideration of the limit as z → 0. As a result, we obtain formulae (20) and (21)^{2}.

Proof 2. Here we use the calculation of the residue. We have

ỹ(z) = y_{n,0}(z) = 1 + c_{1}^{(0)}z + ⋯ + c_{k}^{(0)} z^{k} + ⋯,  z = t^{n−1}. (41)

Therefore,

\frac{c_{k}^{(0)}}{t} + \frac{df_{k}}{dt} = \frac{ỹ(z)}{x^{1+k(n−1)}(1 − x^{(n−1)})^{1+k(n−1)}} ,  (42)

2 Note that other branches of functions y^{(j)}(t) and x^{(j)}(t) ∈ L_{n}, i.e., they are also the linear combinations of the functions u_{1}(z), …, u_{n−1}(z).

or

\frac{c_{k}^{(0)}}{t} + \frac{df_{k}}{dt} = \frac{C_{k}^{(0)}}{x} + \frac{dF_{k}}{dx} .  (43)

Here f_{k}(t) and F_{k}(x) are some rational functions.

The quantity C_{j}^{(0)} is given by the expansion of the function

(1 − z)^{−(1+k(n−1))} = ∑_{j=0}^{∞} C_{j}^{(0)}z^{j}.  (44)

Hence,

c_{k}^{(0)} = C_{k}^{(0)} = ( \begin{matrix}kn \\ k \end{matrix} ) .  (45)

The analogous calculation for the function

y_{n,j}(z) = t^{j} ∑_{k=0}^{∞} c_{k}^{(j)} z^{k}

gives

c_{k}^{(j)} = ( \begin{matrix}kn + j \\ k \end{matrix} ) .

This completes the proof.

From this it follows formula (7) for y_{n,j}(t). Integrating y_{n,j−1}(t) we obtain (21).

So, Theorem 1 is proved.

5. We give here the additional properties of the functions x_{n,j} and y_{n,j}.

Lemma 3. The function y(t) = y_{n,0}(t) satisfies the following nonlinear differential equation

\frac{1}{y} \frac{d}{dt} \frac{1}{y} \frac{d}{dt} ⋯ \frac{1}{y} = − n! .  (46)

Proof. We have

\frac{1}{y} = 1 − nx^{n−1}.  (47)

Differentiating this n − 1 times with respect to t, we obtain equation (46).

Lemma 4. For the function y(t) there is the following formal integral representation^{3}

y(t) = ∫_{ℂ} dμ(z, z̄) exp (z̄(z + t)^{n}) ,  (48)

where

dμ(z, z̄) = \frac{1}{π} exp (−|z|^{2}) dz dz̄.  (49)

Proof. We can check this by expanding of the integrand in the series and using the formula

∫ dμ(z, z̄) z^{j} z̄^{k} = δ_{jk} k! .  (50)

In conclusion, observe that from the evident formulae such as

(x(t))^{j}(x(t))^{l} = (x(t))^{j+l},  (x(t))^{j}(x(t)^{l} y(t)) = (x(t))^{j+l} y(t)  (51)

there follows nontrivial (and probably new) identities for the generalized hypergeometric functions. We give here a couple of them. Lemma 5. The following identities are valid:

nF_{n−1} (\frac{j}{n}, …, \frac{j + n − 1}{n}; \frac{j + 1}{n − 1}, …, \frac{j + n − 1}{n − 1}; z)

× _{n}F_{n−1} (\frac{l}{n}, …, \frac{l + n − 1}{n}; \frac{l + 1}{n − 1}, …, \frac{l + n − 1}{n − 1}; z) =

nF_{n−1} (\frac{j + l}{n}, …, \frac{j + l + n − 1}{n}; \frac{j + l + 1}{n − 1}, …, \frac{j + l + n − 1}{n − 1}; z) ,

(52)

nF_{n−1} (\frac{j}{n}, …, \frac{j + n − 1}{n}; \frac{j + 1}{n − 1}, …, \frac{j + n − 1}{n − 1}; z)

× _{n−1}F_{n−2} (\frac{l + 1}{n}, …, \frac{l + n}{n}; \frac{l + 1}{n − 1}, …, \frac{l + n − 1}{n − 1}; z) =

n−1F_{n−2} (\frac{j + l}{n}, …, \frac{j + l + n}{n}; \frac{l + l + 1}{n − 1}, …, \frac{j + l + n − 1}{n − 1}; z) .

(53)

3 The term formal means that we should first expand the function exp (z̄(z + t)^{n}) in the power series and only then integrate it. Note that for the functions x_{n,j}(t) and y_{n,j}(t)

we also have the standard integral representations (see [12], [13]).

Appendix. Explicit expressions for functions x(t) = x_{n,1}(t) and y_{j}(t) = y_{n,j}(t) for n =2, 3, 4, 5 and 6.

n = 2

x(t) = t_{2}F_{1} (\frac{1}{2}, 1; 2; 4t) ,
x(t) = t + t^{2} + 2 t^{3} + 5 t^{4} + 2 · 7 t^{5} + 2 · 3 · 7 t^{6} + 2^{2} · 3 · 11 t^{7}
+ 3 · 11 · 13 t^{8} + 2 · 5 · 11 · 13 t^{9} + ⋯
y_{0} = _{1}F_{0} (\frac{1}{2}; 4t) ,
y_{0} = 1 + 2 t + 2 · 3 t^{2} + 2^{2} · 5 t^{3}
+ 2 · 5 · 7 t^{4} + 2^{2} · 3^{2} · 7 t^{5} + 2^{2} · 3 · 7 · 11 t^{6}
+ 2^{3} · 3 · 11 · 15 t^{7} + 2 · 3^{2} · 5 · 11 · 13 t^{8} + ⋯
y_{1} = t + 3 t^{2} + 2 · 5 t^{3} + 5 · 7 t^{4} + 2 · 3^{2} · 7 t^{5} + ⋯

n = 3

x(t) = t_{2}F_{1} (\frac{1}{3}, \frac{2}{3}; \frac{3}{2}; \frac{27}{4}t^{2})
x(t) = t + t^{3} + 3 t^{5} + 2^{2} · 3 t^{7} + 5 · 11 t^{9} + 3 · 7 · 13 t^{11}
+ 2^{2} · 3 · 7 · 17 t^{13} + 2^{3} · 3 · 17 · 19 t^{15} + 3^{2} · 11 · 19 · 23 t^{17} + ⋯
y_{0}(t) = _{2}F_{1} (\frac{1}{3}, \frac{2}{3}; \frac{1}{2}; \frac{27}{4}t^{2}) ,
y_{0}(t) = y(t|3, 0) = 1 + 3 t^{2} + 3 · 5 t^{4} + 2^{2} · 3 · 7 t^{6} + 3^{2} · 5 · 11 t^{8}
+ 3 · 7 · 11 · 13 t^{10} + 2^{2} · 3 · 7 · 13 · 17 t^{12}
+ 2^{3} · 3^{2} · 5 · 17 · 19 t^{14} + 3^{2} · 11 · 17 · 19 · 23 t^{16} + ⋯
y_{0}^{−1}(t) = 1 − 3 t^{2} − 2 · 3 t^{4} − 3 · 7 t^{6} − 2 · 3^{2} · 5 t^{8} − 3 · 11 · 13 t^{10}
− 2^{3} · 3 · 7 · 13 t^{12} − ⋯

y_{1}(t) = t_{2}F_{1} (\frac{2}{3}, \frac{4}{3}; \frac{3}{2}; \frac{27}{4}t^{2})
y_{1}(t) = t + 2^{2} t^{3} + 3 · 7 t^{5} + 2^{3} · 3 · 5 t^{7} + 5 · 11 · 13 t^{9}
+ 2^{4} · 3 · 7 · 13 t^{11} + 2^{2} · 3 · 7 · 17 · 19 t^{13} + ⋯

y_{2}(t) = t^{2} _{3}F_{2} (1, \frac{4}{3}, \frac{5}{3}; \frac{3}{2}, 2; \frac{3^{3}}{2^{2}} t^{2})
y_{2}(t) = t^{2} + 5 t^{4} + 2^{2} · 7 t^{6} + 3 · 5 · 11 t^{8}
+ 7 · 11 · 13 t^{10} + ⋯

n = 4

x(t) = t_{3}F_{2} (\frac{1}{4}, \frac{1}{2}, \frac{3}{4}; \frac{2}{3}, \frac{4}{3}; \frac{4^{4}}{3^{3}} t^{3})
x(t) = t + t^{4} + 2^{2} t^{7} + 2 · 11 t^{10} + 2^{2} · 5 · 7 t^{13}
+ 3 · 17 · 19 t^{16} + 2^{2} · 7 · 11 · 23 t^{19} + 2^{2} · 3^{2} · 5 · 13 · 23 t^{22}
+ 2^{2} · 3^{2} · 13 · 29 · 31 t^{25} + ⋯

y_{0}(t) = _{3}F_{2} (\frac{1}{4}, \frac{1}{2}, \frac{3}{4}; \frac{1}{3}, \frac{2}{3}; \frac{4^{4}}{3^{3}} t^{3})
y_{0}(t) = 1 + 2^{2} t^{3} + 2^{2} · 7 t^{6} + 2^{2} · 5 · 11 t^{9} + 2^{2} · 5 · 7 · 13 t^{12}
+ 2^{4} · 3 · 17 · 19 t^{15} + 2^{2} · 7 · 11 · 19 · 23 t^{18}
+ 2^{3} · 3^{2} · 5 · 11 · 13 · 23 t^{21} + 2^{2} · 3^{2} · 5^{2} · 13 · 29 · 31 t^{24} + ⋯

y_{1}(t) = t_{3}F_{2} (\frac{1}{2}, \frac{3}{4}, \frac{5}{4}; \frac{2}{3}, \frac{4}{3}; \frac{4^{4}}{3^{3}} t^{3})
y_{1}(t) = t + 5 t^{4} + 2^{2} · 3^{2} t^{7} + 2 · 11 · 13 t^{10}
+ 2^{2} · 5 · 7 · 17 t^{13} + ⋯

y_{2}(t) = t^{2} _{3}F_{2} (\frac{3}{4}, \frac{5}{4}, \frac{3}{2}; \frac{4}{3}, \frac{5}{3}; \frac{4^{4}}{3^{3}} t^{3})
y_{2}(t) = t^{2} + 2 · 3 t^{5} + 3^{2} · 5 t^{8} + 2^{2} · 7 · 13 t^{11}
+ 2^{2} · 3^{2} · 5 · 17 t^{14} + ⋯

y_{3}(t) = t^{3} _{4}F_{3} (1, \frac{5}{4}, \frac{3}{2}, \frac{7}{4}; \frac{4}{3}, \frac{5}{3}, 2; \frac{4^{4}}{3^{3}} t^{3})
y_{3}(t) = t^{3} + 7 t^{6} + 5 · 11 t^{9} + 5 · 7 · 13 t^{12}
+ 2^{2} · 3 · 17 · 19 t^{15} + ⋯

n = 5

x(t) = t_{4}F_{3} (\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}; \frac{1}{2}, \frac{3}{4}, \frac{5}{4}; \frac{5^{5}}{4^{4}} t^{4})
x(t) = t + t^{5} + 5 t^{9} + 5 · 7 t^{13} + 3 · 5 · 19 t^{17} + 2 · 5 · 11 · 23 t^{21}
+ 3^{2} · 7 · 13 · 29 t^{25} + 2^{3} · 5 · 11 · 17 · 31 t^{29}
+ 3 · 5 · 13 · 17 · 19 · 37 t^{33} + ⋯

y_{0}(t) = _{4}F_{3} (\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}; \frac{1}{4}, \frac{1}{2}, \frac{3}{4}; \frac{5^{5}}{4^{4}} t^{4})
y_{0}(t) = 1 + 5 t^{4} + 3^{2} · 5 t^{8} + 5 · 7 · 13 t^{12} + 3 · 5 · 17 · 19 t^{16}
+ 2 · 3 · 5 · 7 · 11 · 23 t^{20} + 3^{2} · 5^{2} · 7 · 13 · 29 t^{24}
+ 2^{3} · 5 · 11 · 17 · 29 · 31 t^{28}
+ 3^{2} · 5 · 11 · 13 · 17 · 19 · 37 t^{32} + ⋯

y_{1}(t) = t_{4}F_{3} (\frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{6}{5}; \frac{1}{2}, \frac{3}{4}, \frac{5}{4}; \frac{5^{5}}{4^{4}} t^{4})
y_{1}(t) = t + 2 · 3 t^{5} + 5 · 11 t^{9} + 2^{4} · 5 · 7 t^{13}
+ 3^{2} · 5 · 7 · 19 t^{17} + ⋯

y_{2}(t) = t^{2} _{4}F_{3} (\frac{3}{5}, \frac{4}{5}, \frac{6}{5}, \frac{7}{5}; \frac{3}{4}, \frac{5}{4}, \frac{3}{2}; \frac{5^{5}}{4^{4}} t^{4})
y_{2}(t) = t^{2} + 7 t^{6} + 2 · 3 · 11 t^{10} + 2^{3} · 5 · 17 t^{14}
+ 5 · 7 · 11 · 19 t^{18} + 2 · 3^{3} · 5 · 13 · 23 t^{22} + ⋯

y_{3}(t) = t^{3} _{4}F_{3} (\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}; \frac{5}{4}, \frac{3}{2}, \frac{7}{4}; \frac{5^{5}}{4^{4}} t^{4})
y_{3}(t) = t^{3} + 2^{3} t^{7} + 2 · 3 · 13 t^{11} + 2^{4} · 3 · 17 t^{15}
+ 5 · 7 · 11 · 23 t^{19} + 2^{3} · 3^{3} · 5 · 7 · 13 t^{23} + ⋯

y_{4}(t) = t^{4} _{5}F_{4} (1, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \frac{9}{5}; \frac{5}{4}, \frac{3}{2}, \frac{7}{4}; 2; \frac{5^{5}}{4^{4}} t^{4})
y_{4}(t) = t^{4} + 3^{2} t^{8} + 7 · 13 t^{12} + 3 · 17 · 19 t^{16}
+ 2 · 3 · 7 · 11 · 23 t^{20} + 3^{2} · 5 · 7 · 13 · 29 t^{24} + ⋯

n = 6

x(t) = t_{5}F_{4} (\frac{1}{6}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{5}{6}; \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{6}{5}; \frac{6^{6}}{5^{5}} t^{5})
x(t) = t + t^{6} + 2 · 3 t^{11} + 3 · 17 t^{16} + 2 · 11 · 23 t^{21}
+ 3^{3} · 7 · 29 t^{26} + 2^{4} · 3 · 7 · 11 · 17 t^{31}
+ 2 · 13 · 19 · 37 · 41 t^{36} + 2 · 3^{2} · 11 · 23 · 43 · 47 t^{41} + ⋯

y_{0}(t) = _{5}F_{4} (\frac{1}{6}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{5}{6}; \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}; \frac{6^{6}}{5^{5}} t^{6})
y_{0}(t) = 1 + 2 · 3 t^{5} + 2 · 3 · 11 t^{10} + 2^{4} · 3 · 17 t^{15}
+ 2 · 3 · 7 · 11 · 23 t^{20} + 2 · 3^{3} · 7 · 13 · 29 t^{25}
+ 2^{4} · 3 · 7 · 11 · 17 · 31 t^{30} + 2^{3} · 3^{2} · 13 · 19 · 37 · 41 t^{35}
+ 2 · 3^{2} · 11 · 23 · 41 · 43 · 47 t^{40} + ⋯

Acknowledgments. I am grateful to Professors F. Hirzebruch, D. Leites, A. Levine and Yu.I. Manin for useful remarks, Professor M.L. Glasser for the information about his paper and the Max-Planck-Institute für Mathematik, Bonn for hospitality.

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